How do you find the solution to cottheta+8=3cottheta+2 if 0<=theta<2pi?

1 Answer
Jan 10, 2017

theta in {0.322,3.463} (approx.)

Explanation:

color(blue)(cot(theta))+8=3color(blue)(cot(theta))+2

rarr 2color(blue)(cot(theta))=6

rarr color(blue)(cot(theta))=3

If you have a calculator that evaluates arccos directly then
the primary value for color(red)(theta) = color(red)(arccos(3))
My calculator won't do that so I needed to use
color(white)("XX")color(green)(tan(theta)=1/(cot(theta))=1/3)
and then
color(white)("XX")color(red)(theta)=color(red)(arctan(1/3))
for the primary value of color(red)(theta)

This gave: color(red)(theta~~0.321750554)
as the primary value (in Quadrant 1).
Based on CAST notation for the 4 quadrants we know that a secondary value for color(red)(theta) will also exist in Quadrant 3 at
color(white)("XX")color(red)(theta)~~color(red)(0.32175055+pi)=color(red)(3.463343208)